Problem: The lifespans of lions in a particular zoo are normally distributed. The average lion lives $10$ years; the standard deviation is $1.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lion living between $7.2$ and $11.4$ years.
Solution: $10$ $8.6$ $11.4$ $7.2$ $12.8$ $5.8$ $14.2$ $95\%$ $68\%$ $13.5\%$ $13.5\%$ We know the lifespans are normally distributed with an average lifespan of $10$ years. We know the standard deviation is $1.4$ years, so one standard deviation below the mean is $8.6$ years and one standard deviation above the mean is $11.4$ years. Two standard deviations below the mean is $7.2$ years and two standard deviations above the mean is $12.8$ years. Three standard deviations below the mean is $5.8$ years and three standard deviations above the mean is $14.2$ years. We are interested in the probability of a lion living between $7.2$ and $11.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the lions will have lifespans within 2 standard deviations of the average lifespan. It also tells us that $68\%$ of the lions will have lifespans within 1 standard deviation of the mean. The probability of a particular lion living between $7.2$ and $11.4$ years is $\color{orange}{13.5\%} + {68\%}$, or $81.5\%$.